Question

If a normal chord at point t on the parabola y^{2} =4ax subtends a right angle at vertex ,then prove that t=±\sqrt{2}

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Answer 1

Answer:

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Answer 2

Answer:

Proof:  y^{2} =4ax subtends a right angle at vertex ,then  t=±√2

Step-by-step explanation:

REFER FIGURE:

GIVEN:

• Equation of the parabola is $y^{2}=4 a x \ldots (1)$

• Equation of the normal at is:

$\begin{aligned}&t x+y=2 a t+a t^{3} \\&(t x+y) / (2 a t+a t^{3})=1 \ldots \ldots(2)\end{aligned}$

• Homogenise (1) with the help of (2) combined equation of A Q and A R be

$\begin{aligned}&y^{2}=\frac{4 a x(t x+y)}{a\left(2 t+t^{3}\right)} \\&y^{2}\left(2 t+t^{3}\right)=4 t x^{2}+4 x y \\&4 t x^{2}+4 x y-\left(2 t+t^{3}\right) y^{2}=0\end{aligned} \mathrm{AQ}, \mathrm{AR} are perpendicularCoefficient of x^{2} and coefficient of y^{2}=0 \begin{aligned}&4 t-2 t-t^{3}=0 \\&2 t-t^{3}=0\end{aligned}$

                                                $2t=t^3\\2=t^2\\t=+\sqrt{2} or -\sqrt{2}$

y² =4ax subtends a right angle at vertex ,then  t=±√2